Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x+4y &= 8 \\ -3x-4y &= 1\end{align*}$
Answer: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-4y = 3x+1$ Divide both sides by $-4$ to isolate $y$ $y = {-\dfrac{3}{4}x - \dfrac{1}{4}}$ Substitute this expression for $y$ in the first equation. $8x+4({-\dfrac{3}{4}x - \dfrac{1}{4}}) = 8$ $8x - 3x - 1 = 8$ Simplify by combining terms, then solve for $x$ $5x - 1 = 8$ $5x = 9$ $x = \dfrac{9}{5}$ Substitute $\dfrac{9}{5}$ for $x$ back into the top equation. $8( \dfrac{9}{5})+4y = 8$ $\dfrac{72}{5}+4y = 8$ $4y = -\dfrac{32}{5}$ $y = -\dfrac{8}{5}$ The solution is $\enspace x = \dfrac{9}{5}, \enspace y = -\dfrac{8}{5}$.